CENTER OF MASS AND MOMENT OF INERTIA 143
determine x only. Taking a strip of width dx for the element of mass we have
dm — cr 2 y dx
= 2 cr V2 px dx,
where cr is the mass per unit area. Therefore substituting this expression of dm in equation (I) and changing the limits of integration we obtain
2<j f xVZpxdx
- JQ
X =
rx$
C
JQ
2. Find the center of mass of the lamina bounded by the curves ?/2 = 4 ax and y — bx, Fig. 75. Let dx dy be the area of the element of mass, then
dm — a dx dy.
Therefore substituting in equation (I) and introducing the proper nmits of integration we obtain
/»62 s*2'Vrax /»&2 rz'
_ Jo Jbx*dydx . _ J0 Jbx
_ 0
y =
#
/•62 /»2Vox /»62
Jo J&* ^d:C Jo
- 6x) dx f62 (2 VoS - bx) dx
8a 2a,
:562' 6 *